If k = c, then the roots of the equation are:

Correct Answer is : 1 and ca+bc/a+ba+bc

Correct Answer is : 1 and ca+bc/a+ba+bc

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UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 36-37

Consider the following for the next items that follow:

A quadratic equation is given by (a + b) x2 - (a + b + c) x +

k = 0

, where a, b, c are real.

Question : 37 of 120

Marks: +1, -0

k = c

, then the roots of the equation are:

Solution:

Concept:

For a quadratic equation

px^2 + qx + r = 0

, the roots are given by

x = \frac{-q \pm \sqrt{D}}{2p}

, where

D = q^2 - 4pr

Explanation:

Substitute

k = c

in the given equation

(a + b)x^2 - (a + b + c)x + k = 0

This becomes

(a + b)x^2 - (a + b + c)x + c = 0

Here

p = a + b

q = -(a + b + c)

r = c

Compute the discriminant

D

D = [-(a + b + c)]^2 - 4(a + b)(c)

= (a + b + c)^2 - 4ac - 4bc

= a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 4ac - 4bc

= a^2 + b^2 + c^2 + 2ab - 2ac - 2bc

= a^2 + b^2 + (-c)^2 + 2ab + 2a(-c) + 2b(-c)

= (a + b - c)^2

Thus

D = (a + b - c)^2

Now apply the quadratic formula:

x = \frac{(a + b + c) \pm \sqrt{(a + b - c)^2}}{2(a + b)}

x = \frac{(a + b + c) \pm (a + b - c)}{2(a + b)}

Taking the positive sign:

x = \frac{a + b + c + a + b - c}{2(a + b)} = \frac{2(a + b)}{2(a + b)} = 1

Taking the negative sign:

x = \frac{a + b + c - a - b + c}{2(a + b)} = \frac{2c}{2(a + b)} = \frac{c}{a + b}

Hence the roots are

1

and

\frac{c}{a + b}

Answer:

C. 1 and

\frac{c}{a+b}

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